Question:
If \[ a^2-2a-1=0 \] find:
\[ a^2+\frac{1}{a^2} \]
Solution:
Given:
\[ a^2-2a-1=0 \]
\[ a^2-2a=1 \]
Dividing both sides by \[ a \]
\[ a-2-\frac{1}{a}=0 \]
\[ a-\frac{1}{a}=2 \]
Using identity:
\[ \left(a-\frac{1}{a}\right)^2 = a^2+\frac{1}{a^2}-2 \]
Substituting:
\[ (2)^2 = a^2+\frac{1}{a^2}-2 \]
\[ 4 = a^2+\frac{1}{a^2}-2 \]
\[ a^2+\frac{1}{a^2} = 4+2 \]
\[ =6 \]